# Interval of convergence

1. Jul 14, 2013

### Jbreezy

1. The problem statement, all variables and given/known data

Say that you were using ration test for
$\sum_{n=1}^\infty\frac{(-1)^{n+1} (x-4)^n}{n9^n!}\$

2. Relevant equations

3. The attempt at a solution
You take the limit of the above you will get

$\frac {1}{9} |x-4|$

Book says radius of convergence is 9? Is this done by using the ratio test and making an inequality?

$\frac {1}{9} |x-4| <1$
$|x-4| < 9$
No?

OK then the book says in the next line the

$-5< x< 13$
How did they get this? I thought that they maybe said if R = 9 then put ,-9 and 9 in the inequality. So,
$|(-9)-4| = -13$ and $|(9)-4| = 5$
But that doesn't work because it is backwards theirs was -5 and 13 not 13 and -5
What is going on. If you can elaborate the idea of radius of convergence and how they determined the interval please!

2. Jul 14, 2013

### Jbreezy

HA! nevermind

3. Jul 15, 2013

### HallsofIvy

Staff Emeritus
For others who might be wondering, yes, this is correct. The "ratio test" says that a series , $\sum a_n$ converges (absolutely) as long as
$$\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|< 1$$

Here, that ratio is $[|x- 4|^{n+1}]/[(n+1)9^{n+1}][n 9^n]/[|x- 4|^n= [n/(n+1)]|x- 4|/9$. (I have ignored the "$(-1)^{n+1}$" because of the absolute value.)

The limit as n goes to infinity is, as Jbreezy said, |x- 4|/9. From |x- 4|/9< 1 we obviously get |x- 4|< 9 and so -9< x- 4< 9. The radius of convergence, the distance from the center point, x= 4. to each end of the interval of convergence is 9. So, adding 4 to each part,-9+ 4= -5< x< 13= 9+ 4.
Your "error" is trying to incorporate the absolute value where it does not belong.

Now, the series might or might (or might converge but not "absolutely") at x= -5 and x= 13. Those you would have to check separately.